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Topic: N2 lone pairs (MO Theory)  (Read 5123 times)

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Offline owk9688

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N2 lone pairs (MO Theory)
« on: June 20, 2017, 04:50:43 PM »
I've been playing around with qualitative MO theory recently in preparation for a Physical Organic cumulative exam coming up. For one of the simpler molecules, N2, The complete MO diagram predicts all electrons in either bonding or antibonding orbitals; there are no "nonbonding" combinations that would lead to lone pairs. However, I can see that the two lowest energy levels are a 2s sigma and a 2s sigma star, which essenially cancel, So I guess my question is this:

Are the 2 lone pairs in Nitrogen in 2s orbitals (arising from the 2s of one and the 2s of another not mixing because doing so results in no net stability). If not, are there any lone pairs at all, or are all electrons in either bonding or antibonding orbitals like the M diagram predicts?

Thank you

Offline phth

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Re: N2 lone pairs (MO Theory)
« Reply #1 on: June 21, 2017, 01:01:55 AM »
The two lone pairs in the 2s orbitals are non-bonding because there are electrons in the antibonding ortbial.  There cannot be a bond that has electrons in the antibonding orbital because there is a 0 % probability of finding an electron between the orbitals. This is called that alpha effect: e.g. H2O2. In short, it doesn't matter if there are 0, 1, 2, 3, or 4 electrons in two orbitals because the wave functions will always add together and one orbital will be lower in energy ibid one higher in energy. Non-bonding is kind of a hypothetical thing.

Offline Enthalpy

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Re: N2 lone pairs (MO Theory)
« Reply #2 on: June 22, 2017, 03:49:24 PM »
[...] not mixing because doing so results in no net stability [...]
Orbitals mix as soon as the nuclei as near enough to an other (where "near enough" is a fuzzy limit). The resulting MO are occupied in the case you cite. If, in a first analysis (...not in a more detailed one), the shifts of energy in the bonding and antibonding MO cancel out, the mixing (here of 2s) has little effect on the N≡N bond strength, but it is still observable in the molecule's light absorption spectrum.

I vaguely imagine - with no hard evidence! - that under the high pressure of bubble implosion, atoms (for instance of argon) are brought to create metastable bonds, and after the density has dropped, the total electronic energy is higher than for separated argon atoms. The de-excitation would then explain sonoluminescence, in a process resembling excimer lamps. Some authors claim that gas temperature resulting from compression suffice to explain the emitted light, others claim it doesn't.

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